Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 2x}{x - 8} = \dfrac{x + 40}{x - 8}$
Explanation: Multiply both sides by $x - 8$ $ \dfrac{x^2 - 2x}{x - 8} (x - 8) = \dfrac{x + 40}{x - 8} (x - 8)$ $ x^2 - 2x = x + 40$ Subtract $x + 40$ from both sides: $ x^2 - 2x - (x + 40) = x + 40 - (x + 40)$ $ x^2 - 2x - x - 40 = 0$ $ x^2 - 3x - 40 = 0$ Factor the expression: $ (x - 8)(x + 5) = 0$ Therefore $x = 8$ or $x = -5$ However, the original expression is undefined when $x = 8$. Therefore, the only solution is $x = -5$.